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MATHMAA

# 17thKVS-JMO2014-Solutions

## 17thKVS-JMO2014-Solutions

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1Q) Prove that for no integer n, $$n^{6}$$ + 3 $$n^{5}$$ -5 $$n^{4}$$ - 15 $$n^{3}$$ + 4 $$n^{2}$$ + 12n + 3 is a perfect square .

Solution :  We have to prove that there is no integer n, such that $$n^{6}$$ + 3 $$n^{5}$$ -5 $$n^{4}$$ - 15 $$n^{3}$$ + 4 $$n^{2}$$ + 12n + 3 is a perfect square.

Let us factorize it leaving  constant term as it is.

$$n^{6}$$ + 3 $$n^{5}$$ -5 $$n^{4}$$ - 15 $$n^{3}$$ + 4 $$n^{2}$$ + 12n + 3 = $$n^{5}$$(n+3)-5$$n^{3}$$(n+3) + 4n(n+3) +3

= (n+3)$$(n^{5}-5n^{3}+4n)$$ + 3

=(n+3)n($$n^{4}-5n^{2}+4)$$ +3

=n(n+3)$$(n^{2}-4)(n^{2}-1)$$+3

=n(n+3)(n+2)(n-2)(n+1)(n-1)+ 3

= (n-2)(n-1)n(n+1)(n+2)(n+3) +3  ............(1)

We know that product of six consecutive terms is divisible by 6!=720

(n-2)(n-1)n(n+1)(n+2)(n+3) = 720k, where k is any integer.

We rewrite the equation (1) as

720k+3 =3(240k+1)

240k+1 is not divisible by 3 as 1 is not divisible by 3.

Therefore 3(240k+1) is not a perfect square for any integer.

Hence there is no integer 'n' such that
$$n^{6}$$ + 3 $$n^{5}$$ -5 $$n^{4}$$ - 15 $$n^{3}$$ + 4 $$n^{2}$$ + 12n + 3 is a perfect square.

_________________________ 17thKVS-JMO2014-Solutions _____________________

_________________________ 17thKVS-JMO2014-Solutions _____________________

2Q) Two dice are thrown once simultaneously. Let E be the event "Sum of numbers appearing on the dice ." what are the members of E ? Can you load these dice(not necessarily in the same way) such that all members of E are equally likely ? Give justification.

Solution :

We know that each die has numbers 1 , 2 , 3 , 4 , 5 or 6 . Means each die has 6 events.

If we throw two dice we get 6*6 =36 events. They are

(1, 1) , (1, 2) , (1, 3) , (1, 4) , (1, 5) , (1, 6)

(2, 1) , (2, 2) , (2, 3) ,  (2, 4) , (2, 5) , (2, 6)

(3, 1) , (3, 2) , (3, 3) , (3, 4) , (3, 5) , (3, 6)

(4, 1) , (4, 2) , (4, 3) , (4, 4) , (4, 5) , (4, 6)

(5, 1) , (5, 2) , (5, 3) , (5, 4) , (5, 5) , (5, 6)

(6, 1) , (6, 2) , (6, 3) , (6, 4) , (6, 5) , (6, 6) .

E be the event  " Sum of the numbers appear on the dice"

We  know that minimum sum of the numbers on the dice = 1+1=2

and maximum sum =6+6 =12

So 2, 3, 4, 5, 6, 7, 8, 9, 10 , 11 , 12  (sum of the numbers on dice) are members of E .

E = {2, 3, 4, 5, 6, 7,  8 , 9 , 10 , 11 , 12 }

Now let us find the probability of each event of E.

Chances of getting 2 are (1,1) that is only one possible way.
Probability of getting 2 is 1/36             ($$\because$$ Total number of events are 36)
That is, P(2) = 1/36

Similarly,
P(3)   = 2/36   as chances of getting 3  are (1,2) and (2,1).
P(4)   = 3/36   as chances of getting 4  are (1,3) (2,2) and (3,1).
P(5)   = 4/36   as chances of getting 5  are (1,4) (2,3) (3,2) and (4,1).
P(6)   = 5/36   as chances of getting 6  are (1,5) (2,4) (3,3) (4,2) and (5,1).
P(7)   = 6/36   as chances of getting 7  are (1,6) (2,5) (3,4) (4,3) (5,2) and (6,1).
P(8)   = 5/36
as chances of getting 8  are (2,6) (3,5) (4,4) (5,3) and (6,2).
P(9)   = 4/36
as chances of getting 9  are (3,6) (4,5) (5,4) and (6,3).
P(10) = 3/36
as chances of getting 10 are (4,6) (5,5) and (6,4).
P(11) = 2/36
as chances of getting 11 are (5,6) and (6,5).
P(12) = 1/36
as chances of getting 12 is (6,6).

Now Let us find the events such that members of events are equally likely events.

E$$_{1}$$ = { 2, 12} here P(2)=P(12)=1/36 . Similarly we get ,

E$$_{2}$$ = {3, 11}

E$$_{3}$$ ={4, 10}

E$$_{4}$$ = {5, 9}

E$$_{5}$$ = {6, 8}

E$$_{6}$$ = {7}

Therefore E$$_{1}$$, E$$_{2}$$, E$$_{3}$$, E$$_{4}$$, E$$_{5}$$, E$$_{6}$$ are events such that their members are equally likely events.